This document summarizes the design of a concrete frame structure with a span of 56 inches and a height of 48 inches. It includes calculations for flexural design of beams, modeling of the frame using truss analogies, checking stress requirements in nodal zones, and reinforcement design of beams, ties, and struts. Reinforcement sizes and spacing are determined to satisfy strength and serviceability limits. Diagrams illustrate the truss model and nodal zones analyzed.

1. BACH. RONALD J. PURCA

2. Resistencia Concreto f’c=4000 psi
Resistencia de Fluencia del Acero fy=60000 psi
𝑎 𝑣
ℎ
=
56𝑖𝑛
48𝑖𝑛
= 1.17 < 2 (𝑉𝑖𝑔𝑎 𝑎𝑙𝑡𝑎)

3. 2.1 Diseño a Flexión
d = 44.4in
𝑎 =
𝐴 𝑠 𝑓𝑦
0.85𝑓′ 𝑐 𝑏
𝑀 𝑛 = 𝐴 𝑠 𝑓𝑦 𝑑 −
𝑎
2
𝑀 𝑢 = 11984 𝑘𝑙𝑏. 𝑖𝑛 ≤ 0.9𝑀 𝑛
𝐴 𝑠 ≥ 5.4 𝑖𝑛2
𝑐 ≥
𝑎
0.85
≈ 8.1 𝑖𝑛

4. Modelo Reticulado: Bielas (Azul)
Tirantes (Rojo)
1
6
2
71
2
3
4
3 5 4

5. Modelo Reticulado: Bielas (Azul)
Tirantes (Rojo)
307 klb (C)
307 klb (T)
154 klb (C)
154 klb (T)
214klb(T)
214 klb 214 klb

6. Los modelos de bielas y tirantes fallan debido a:
o Aplastamientos de las bielas (extremos)
o Aplastamiento en zonas nodales (caras)
o Fluencia en los tirantes
o Falta de anclaje de tirantes

7. ZONA NODAL 1 CCT
1
3
263 Klb
154 Klb
214 Klb
𝑤𝑟𝑒𝑞 =
𝐹𝑢
𝛽235.7
=
154
(0.80)(35.7)
= 5.4𝑖𝑛
8.0 in > 5.4 in OK
𝑤𝑟𝑒𝑞 =
𝐹𝑢
𝛽235.7
=
214
(0.80)(35.7)
= 7.5𝑖𝑛
16 in > 7.5 in OK
𝑤𝑟𝑒𝑞 =
𝐹𝑢
𝛽235.7
=
263
(0.80)(35.7)
= 9.2𝑖𝑛
17.7 in > 9.2 in OK
6

8. ZONA NODAL 2 CCT
2
2
3
263 Klb
154 Klb
214 Klb
𝑤𝑟𝑒𝑞 =
𝐹𝑢
𝛽235.7
=
154
(0.80)(35.7)
= 5.4𝑖𝑛
10 in > 5.4 in OK
𝑤𝑟𝑒𝑞 =
𝐹𝑢
𝛽235.7
=
214
(0.80)(35.7)
= 7.5𝑖𝑛
14.6 in > 7.5 in OK
𝑤𝑟𝑒𝑞 =
𝐹𝑢
𝛽235.7
=
263
(0.80)(35.7)
= 9.2𝑖𝑛
17.7 in > 9.2 in OK

9. ZONA NODAL 4 CCCC
4
2
4
263 Klb 307 Klb
214 Klb
𝑤𝑟𝑒𝑞 =
𝐹𝑢
𝛽235.7
=
307
(1)(35.7)
= 8.6𝑖𝑛
10 in > 8.6 in OK
𝑤𝑟𝑒𝑞 =
𝑉𝑢
𝛽235.7
=
214
(1)(35.7)
= 6.0𝑖𝑛
16 in > 6.0 in OK
𝑤𝑟𝑒𝑞 =
𝐹𝑢
𝛽235.7
=
263
(1)(35.7)
= 7.4𝑖𝑛
18.8 in > 7.4 in OK
1
154 Klb
𝑤𝑟𝑒𝑞 =
𝐹𝑢
𝛽235.7
=
154
(1)(35.7)
= 4.3𝑖𝑛
10 in > 4.3 in OK

10. ZONA NODAL 3 CTTT
3 7
6
263 Klb
307 Klb
214 Klb
𝑤𝑟𝑒𝑞 =
𝐹𝑢
𝛽235.7
=
307
(0.6)(35.7)
= 14.3𝑖𝑛
8 in < 14.3 in NO PASA
𝑤𝑟𝑒𝑞 =
𝑉𝑢
𝛽235.7
=
214
(0.6)(35.7)
= 10𝑖𝑛
23.2 in > 10 in OK
𝑤𝑟𝑒𝑞 =
𝐹𝑢
𝛽235.7
=
263
(0.6)(35.7)
= 12.3𝑖𝑛
18.8 in > 12.3 in OK
4
154 Klb
𝑤𝑟𝑒𝑞 =
𝐹𝑢
𝛽235.7
=
154
(0.6)(35.7)
= 7.2𝑖𝑛
8 in > 7.2 in OK

11. Ancho de biela 3 = 17.7in (Provisto) > 9.9in OK
𝑤𝑟𝑒𝑞 =
𝐹𝑢
∅ 0.85 𝛽2 𝑓′ 𝑐 𝑏
=
263000 𝑙𝑏𝑓
0.75 0.85 (0.75)(4000𝑝𝑠𝑖)(14𝑖𝑛)
= 9.9𝑖𝑛
1
3
BIELA 3 (Nodo 1)

12. Ancho de biela 3 = 17.7in (Provisto) > 9.9in OK
𝑤𝑟𝑒𝑞 =
𝐹𝑢
∅ 0.85 𝛽2 𝑓′ 𝑐 𝑏
=
263000 𝑙𝑏𝑓
0.75 0.85 (0.75)(4000𝑝𝑠𝑖)(14𝑖𝑛)
= 9.9𝑖𝑛
2
3
BIELA 3 (Nodo 2)

13. 𝑤𝑟𝑒𝑞 =
𝐹𝑢
𝛽235.7
=
154
(1)(35.7)
= 4.3𝑖𝑛
BIELA 1
10.0 in > 4.3 in OK
𝑤𝑟𝑒𝑞 =
𝐹𝑢
𝛽235.7
=
307
(1)(35.7)
= 8.6𝑖𝑛
BIELA 2
10.0 in > 8.6 in OK

14. 𝑤𝑟𝑒𝑞 =
𝐹𝑢
𝛽235.7
=
263
(0.75)(35.7)
= 9.9𝑖𝑛
BIELA 4
18.8 in > 9.9 in OK

15. TIRANTE 6
𝐴 𝑠 =
𝐹𝑢
0.75𝑓𝑦
𝐴 𝑠 =
154000 𝑙𝑏𝑓
(0.75)(60000𝑝𝑠𝑖)
= 3.4𝑖𝑛2
𝐿 𝑑ℎ =
0.02 (1)(1)(60000𝑝𝑠𝑖)
4000𝑝𝑠𝑖
= 19𝑖𝑛
6 Varillas #8 (4.74in2) en 2 capas
Si recubrimiento > 2.5in  Ldh=0.70(19in)= 13.3in
16+4/tan(54.3°)-1.5-0.625= 16.7in

16. TIRANTE 7
𝐴 𝑠 =
𝐹𝑢
0.75𝑓𝑦
𝐴 𝑠 =
307000 𝑙𝑏𝑓
(0.75)(60000𝑝𝑠𝑖)
= 6.8𝑖𝑛2
𝐿 𝑑ℎ =
60000𝑝𝑠𝑖 (1)(1)(1)(1𝑖𝑛)
25 4000𝑝𝑠𝑖
= 38𝑖𝑛
6 Varillas #8 + 2 #8 + 2#6 (7.2in^2) 3 Capas
𝐿 𝑑ℎ =
60000𝑝𝑠𝑖 (1)(1)(1)(3/4𝑖𝑛)
25 4000𝑝𝑠𝑖
= 28.5𝑖𝑛
36-1.5-0.625= 33.9in

17. TIRANTE 5
𝐴 𝑠 =
𝐹𝑢
0.75𝑓𝑦
𝐴 𝑠 =
214000 𝑙𝑏𝑓
(0.75)(60000𝑝𝑠𝑖)
= 4.8𝑖𝑛2
1 Varilla #5 (0.31 in2)
Luego: 4.8/ 0.31 /2 = 7.75 Usar 8 estribos cerrados #5
𝑠 =
𝐴 𝑣
0.0025𝑏
=
(2)0.31
0.0025(14)
= 17.7𝑖𝑛
Usar estribos cerrados #5 @ 6in

18. 𝑠 𝑣 =
𝐴 𝑠 𝑠𝑒𝑛(∝ °)
0.003𝑏𝑠
=
(2)(0.2)𝑠𝑒𝑛(54.3°)
0.003(14)
= 7.6𝑖𝑛
1 Varilla #4 (0.2 in2)
Usar Varillas horizontales #4 @ 7in